If a massive sphere spins faster and faster, will a black hole appear?

Question

If we make an object moving faster and faster in 1 direction no black hole will not be produced. We can always find a frame in which the mass is at rest. Even though we have put a lot of work into it, the rest mass will not increase. Only the relative (relativistic) mass increases. This can be used to make a black hole appear if we let the mass meet with another (equal) mass in a frame that is at rest wrt to the frame of the speeding mass. Meaning that the increase in mass is indeed relative.
But what happens if we make a massive object rotate with increasing speed? We put more and more energy in it so the relative mass increases. The frame in which the mass is at rest is not an inertial frame, as in the linear case. It is a co-rotating frame in which a gravity field is present but the mass is at rest. So you would expect that in this case a black hole is formed without the need to let it meet another mass like in the linear case (the increase in mass is not relative?).
Because the gravitational field of the rotating mass is determined by the (rest)mass and momentum the field will be different from a static mass in the sense that frame-dragging is present. Will this lead to the formation of a black hole? Is the Kerr metric important in this case? I mean, this is the metric of a rotating black hole, of which I want to know if it appears in the first place.
Will the only effect of the spinning be that space is super-dragged? In other words, will there be no change in the radial part of the metric but only in the radial part? Will the radial part show time dependence?

So, in short:
Let's assume that the object is an incompressible sphere, to avoid deformation. Let's ignore the paradoxes of length contraction (or can't we?). How will the spacetime around the sphere evolve if it rotates faster and faster?

One last thing. Suppose we look at the sphere at an instant. It is composed of infinitesimal pieces of mass that all have a linear velocity. Each piece of mass is contracted (that is, the space it occupies) due to length contraction, which will give problems. How can a sphere contract in an angular direction? Will its volume become smaller? But aside from that, can't we say that because linear moving masses do not result in a black hole, a rotating sphere will not either? A linear moving mass will only give rise to frame dragging in the direction of motion, so is this not the case here too? Or can you say that because the rotational kinetic energy of the sphere increases (which is also the case in linear motion, though there no black hole arises because the linear kinetic energy does not stay in the same volume) a black hole must see the daylight? So will it only be frame-dragging or will a hole be there? It must be the last, because there will be kinetic energy present in a "stationary" volume (unlike the linear case) that approaches infinity. Can we say that the sphere, due to special relativistic effects, shrinks to a sphere of zero radius (though mechanical incompressible or inexpandable)? Which results in a black hole? On the other hand, will the centrifugal force not be able to prevent the hole from forming? Suppose a neutron star, as mentioned by @Andrew, was rotating that fast that the gravity force holding it together would be comparable to the centrifugal force (both forces are proportional to $\frac{1}{r}$). The star will never become a hole in this case (the question of how the star got in that state if a different one though). It's complicated indeed...

Answer 1

Wikipedia gives the total gravitational mass-energy sum of a Kerr black hole as:

$$M_{\text{total}} = \sqrt{M_{\text{irr}}^2 + \frac{J^2c^2}{4M_{\text{irr}}^2G^2}},$$

where $M_{\text{total}}$ is the equivalent total gravitational mass and $M_{\text{irr}}$ is the irreducible mass, which is basically the mass when the star is not spinning.

Imagine we have a rapidly spinning neutron star that has an angular momentum that is just sufficient to prevent it collapsing to a Kerr black hole. At this point, the $\dfrac{2GM_{\text{total}}}{Rc^2}$ term in the inverse gravitational gamma factor $$\sqrt{1-\frac{2GM_{\text{total}}}{Rc^2}}$$ is approaching unity. If the term exceeds unity the inverse gravitational gamma factor becomes imaginary and a black hole is formed.

We can imagine firing neutrons at the neutron star tangentially in such a way to increase the angular momentum $J$. The question is, can we increase $J$ without increasing $M_{\text{total}}$? The neutrons we are firing at it have mass, so this increases the $M^2_{\text{irr}}$ term. We can write the increase in the $ \dfrac{J^2 c^2}{4M_{\text{irr}}^2G^2}$ term as $\dfrac{rv^2}{4G^2m}$, where $v$ and $m$ are the tangential velocity and mass of the injected neutron, respectively, and $r$ is the tangential radius. This term will also increase $M_{\text{total}}$ for any $m$ and any $v$ that increase the angular momentum. It is inevitable that increasing the angular momentum of a rapidly spinning neutron star, that is near extremal, will eventually collapse it to a black hole. This is due to the increase in effective gravitational mass energy due to the increased energy when increasing the angular momentum. Even if we can find a way to increase the angular momentum without increasing the mass of the neutron star, the effective total mass will still increase and collapse is inevitable with sufficient increase in spin for an initially sufficiently compact star. We can bring a spinning neutron star to a near extremal state simply by dripping neutrons on it. If we do this in a manner that does not increase the angular momentum or even reduces it, it will eventually approach the extremal state as the mass increases.

P.S. the above does not consider extreme relativistic tangential velocities that approach the speed of light. Transverse force orthogonal to the tangential velocity increases by the gamma factor in Special Relativity. This increases the centrifugal force that must be overcome in order for collapse to occur. It is possible that a neutron star with extreme relativistic tangential velocities might not collapse with increased angular velocity but it is difficult to imagine how to get to such a state without collapse occurring before those extreme angular velocities are reached.

P.P.S. it is generally agreed that in the case of linearly accelerating a gravitational body to a relativistic velocity, the body will not collapse to a black hole, as we can always find another inertial observer that is at rest with body where its relativistic mass has not increased. However, in the rotational case, rotation is absolute and collapse can occur due to increased rotation.

P.P.P.S If we start with a far from extremal body like the Earth or a large gas cloud and start spinning it faster, the surface gravity reduces and there is tendency for the volume to increase and reduce the density of the body. In this case the body will not become a black hole no matter how fast it spins, because there is a minimum density required in order for a black hole to form. The above analysis requires the body to be near extremal in the first place with the mass concentrated in a small enough volume for its density to be close to that required for a black hole, before the rotation is increased.

Answer 2

Since mass increases while object rotates, we can re-state Schwarzschild radius of rotating body:

$$ \tag 1 r_{\text{s}}={\frac {2GM\gamma}{c^{2}}} $$ injecting Lorentz factor into it, since rotating body tangential speed can't overcome light speed $c$, so as body acquires relativistic energy-mass due to rotation, it's Schwarzschild radius increases until it reaches real rotating object radius, $r_s \to r$ and so rotating object becomes a black hole.

From (1) we can infer critical $\gamma$ factor for a rotating object becoming black hole:

$$ \tag 2 {\gamma = \frac {r_{\text{s}}c^{2}}{2GM}}$$

Let's calculate this gamma factor for a massive neutron star PSR J0952–0607 which already rotates with amazing tangential speed about $0.14c$ and has mass $\approx 2.35M_{☉}$, assuming standard neutron star radius of $10~\text{km}$, from (2) we get that in case $\gamma = 1.44$ this neutron star should become a black hole.

As per tangential speed extrapolated from Lorentz factor, $$ \tag 3 \frac vc = \sqrt{ 1-\frac{1}{\gamma^2} } $$

we get that this pulsar must rotate at $0.72c$ tangential speed, or at blazing angular speed of $\approx 206~000~\text{RPM}$, until event horizon catches star current radius.

HTH!